Calculus Derivative Question

Hi Girldesi :)

 

If g(x)=x4-2, find g'(1) and use it to find an equation of the tangent line to the curve y=x^4-2 at the point (1,-1).

 

\(g(x)=x^4-2\\ g'(x)=4x^3\\ g'(1)=4*1^3\\ g'(1)=4\\ \)

 

The thing you must ALWAYS remember is the the first dirivative at a particular point IS the gradient of the tangent to the curve at that point.

so now you just have to find the line with gradient -4 through the point (1,-1)

You should be able to do that yourself but if you can't ask me to show you.

 

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Askers comment:  I dont know how to solve using the equation: f(x+h)-f(x)/h

I have already answered but now I iwill show you hos to find the gradient of the tangent from first principals

 

 

 

\(\begin{align} gradient&=\displaystyle\lim_{h \rightarrow 0}\;\frac{f(x+h)-f(x)}{h}\qquad where\;\;x=1\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{((1+h)^4-2)-(1-2)}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{((1+h)^4-1}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h^4+4h^3+6h^2+4h+1-1}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h^4+4h^3+6h^2+4h}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h(h^3+4h^2+6h+4)}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\: h^3+4h^2+6h+4\\ &=4 \end{align} \)

 

Again you still have to find the equation of the line through (1,-1) with a gradient of 4

 

If you need more explanation then please ask for it :))